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Question Number 65334 by Tawa1 last updated on 28/Jul/19

Answered by mr W last updated on 28/Jul/19

$$\mathrm{\angle }D=360-126-42\times 2=150°$$$$\frac{DC}{\sin x}=\frac{AC}{\sin 150}$$$$\Rightarrow DC=\frac{\sin x}{\sin 150}\times AC$$$$\frac{AC}{\sin 126}=\frac{AB}{\sin [42-(30-x)]}=\frac{AB}{\sin (x+12)}$$$$\Rightarrow AB=\frac{\sin (x+12)}{\sin 126}\times AC$$$$DC=AB$$$$\Rightarrow \frac{\sin x}{\sin 150}=\frac{\sin (x+12)}{\sin 126}$$$$\Rightarrow \frac{\sin x}{\sin 30}=\frac{\sin x\cos 12+\cos x\sin 12}{\sin 54}$$$$\Rightarrow \frac{1}{\sin 30}=\frac{\cos 12+\cot x\sin 12}{\cos 36}$$$$\Rightarrow \cot x=\frac{2\cos 36-\cos 12}{\sin 12}$$$$\Rightarrow \cot x=\frac{(8{\cos }^{2}12-7)\cos 12}{\sin 12}$$$$\Rightarrow \cot x=\frac{1-8{\sin }^{2}12}{\tan 12}$$$$\Rightarrow x=18°$$

Commented by Tawa1 last updated on 28/Jul/19

$$\mathrm{Great}\mathrm{sir},\mathrm{God}\mathrm{bless}\mathrm{you}.\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}.$$

Commented by Tony Lin last updated on 29/Jul/19

$$howtoknow\frac{1-8{sin}^{2}12}{tan12}=cot18$$

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